Fraunhofer Diffraction Circular aperture Wed Nov 27 2002

Fraunhofer Diffraction: Circular aperture Wed. Nov. 27, 2002 1

Fraunhofer diffraction from a circular aperture y x P r Lens plane 2

Fraunhofer diffraction from a circular aperture Do x first – looking down Path length is the same for all rays = ro Why? 3

Fraunhofer diffraction from a circular aperture Do integration along y – looking from the side P +R y=0 -R ro r = ro - ysin 4

Fraunhofer diffraction from a circular aperture Let Then 5

Fraunhofer diffraction from a circular aperture The integral where J 1( ) is the first order Bessell function of the first kind. 6

Fraunhofer diffraction from a circular aperture n These Bessell functions can be represented as polynomials: n and in particular (for p = 1), 7

Fraunhofer diffraction from a circular aperture n Thus, n where = k. Rsin and Io is the intensity when =0 8

Fraunhofer diffraction from a circular aperture n Now the zeros of J 1( ) occur at, = 0, 3. 832, 7. 016, 10. 173, … = 0, 1. 22 , 2. 23 , 3. 24 , … =k. R sin = (2 / ) sin • Thus zero at sin = 1. 22 /D, 2. 23 /D, 3. 24 /D, … 9

Fraunhofer diffraction from a circular aperture The central Airy disc contains 85% of the light 10

Fraunhofer diffraction from a circular aperture D sin = 1. 22 /D 11

Diffraction limited focussing n n sin = 1. 22 /D The width of the Airy disc W = 2 fsin 2 f = 2 f(1. 22 /D) = 2. 4 f /D W = 2. 4(f#) > f# > 1 n Cannot focus any wave to spot with dimensions < f D 12

Fraunhofer diffraction and spatial resolution n Suppose two point sources or objects are far away (e. g. two stars) Imaged with some optical system Two Airy patterns ¨ S 1 If S 1, S 2 are too close together the Airy patterns will overlap and become indistinguishable S 2 13

Fraunhofer diffraction and spatial resolution n Assume S 1, S 2 can just be resolved when maximum of one pattern just falls on minimum (first) of the other Then the angular separation at lens, n e. g. telescope D = 10 cm = 500 X 10 -7 cm n e. g. eye D ~ 1 mm min = 5 X 10 -4 rad n 14

Polarization 15

Matrix treatment of polarization y Ey n Ex x Consider a light ray with an instantaneous Evector as shown 16

Matrix treatment of polarization n Combining the components n The terms in brackets represents the complex amplitude of the plane wave 17

Jones Vectors n The state of polarization of light is determined by the relative amplitudes (Eox, Eoy) and, ¨ the relative phases ( = y - x ) of these components ¨ n The complex amplitude is written as a twoelement matrix, the Jones vector 18

Jones vector: Horizontally polarized light n n The electric field oscillations are only along the x-axis The Jones vector is then written, The arrows indicate the sense of movement as the beam approaches you y x where we have set the phase x = 0, for convenience The normalized form is 19

The electric field. Vertically polarized light Jones vector: n n oscillations are only along the y-axis The Jones vector is then written, y x n Where we have set the phase y = 0, for convenience The normalized form is 20

Jones vector: Linearly polarized light at an arbitrary angle n n If the phases are such that = m for m = 0, 1, 2, 3, … Then we must have, y x and the Jones vector is simply a line inclined at an angle = tan-1(Eoy/Eox) since we can write The normalized form is 21

Jones vector and polarization y Eoy n b In general, the Jones vector for the arbitrary case a x is an ellipse Eox 22